Metamath Proof Explorer


Theorem 2atjlej

Description: Two atoms are different if their join majorizes the join of two different atoms. (Contributed by NM, 4-Jun-2013)

Ref Expression
Hypotheses ps1.l ⊢ ≤ ˙ = ≤ K
ps1.j ⊢ ∨ ˙ = join ⁡ K
ps1.a ⊢ A = Atoms ⁡ K
Assertion 2atjlej ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q ∧ R ∈ A ∧ S ∈ A ∧ P ∨ ˙ Q ≤ ˙ R ∨ ˙ S → R ≠ S

Proof

Step Hyp Ref Expression
1 ps1.l ⊢ ≤ ˙ = ≤ K
2 ps1.j ⊢ ∨ ˙ = join ⁡ K
3 ps1.a ⊢ A = Atoms ⁡ K
4 simp33 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q ∧ R ∈ A ∧ S ∈ A ∧ P ∨ ˙ Q ≤ ˙ R ∨ ˙ S → P ∨ ˙ Q ≤ ˙ R ∨ ˙ S
5 simp1 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q ∧ R ∈ A ∧ S ∈ A ∧ P ∨ ˙ Q ≤ ˙ R ∨ ˙ S → K ∈ HL
6 simp21 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q ∧ R ∈ A ∧ S ∈ A ∧ P ∨ ˙ Q ≤ ˙ R ∨ ˙ S → P ∈ A
7 simp22 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q ∧ R ∈ A ∧ S ∈ A ∧ P ∨ ˙ Q ≤ ˙ R ∨ ˙ S → Q ∈ A
8 simp23 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q ∧ R ∈ A ∧ S ∈ A ∧ P ∨ ˙ Q ≤ ˙ R ∨ ˙ S → P ≠ Q
9 simp31 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q ∧ R ∈ A ∧ S ∈ A ∧ P ∨ ˙ Q ≤ ˙ R ∨ ˙ S → R ∈ A
10 simp32 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q ∧ R ∈ A ∧ S ∈ A ∧ P ∨ ˙ Q ≤ ˙ R ∨ ˙ S → S ∈ A
11 1 2 3 ps-1 ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q ∧ R ∈ A ∧ S ∈ A → P ∨ ˙ Q ≤ ˙ R ∨ ˙ S ↔ P ∨ ˙ Q = R ∨ ˙ S
12 5 6 7 8 9 10 11 syl132anc ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q ∧ R ∈ A ∧ S ∈ A ∧ P ∨ ˙ Q ≤ ˙ R ∨ ˙ S → P ∨ ˙ Q ≤ ˙ R ∨ ˙ S ↔ P ∨ ˙ Q = R ∨ ˙ S
13 4 12 mpbid ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q ∧ R ∈ A ∧ S ∈ A ∧ P ∨ ˙ Q ≤ ˙ R ∨ ˙ S → P ∨ ˙ Q = R ∨ ˙ S
14 2 3 lnnat ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A → P ≠ Q ↔ ¬ P ∨ ˙ Q ∈ A
15 5 6 7 14 syl3anc ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q ∧ R ∈ A ∧ S ∈ A ∧ P ∨ ˙ Q ≤ ˙ R ∨ ˙ S → P ≠ Q ↔ ¬ P ∨ ˙ Q ∈ A
16 8 15 mpbid ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q ∧ R ∈ A ∧ S ∈ A ∧ P ∨ ˙ Q ≤ ˙ R ∨ ˙ S → ¬ P ∨ ˙ Q ∈ A
17 13 16 eqneltrrd ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q ∧ R ∈ A ∧ S ∈ A ∧ P ∨ ˙ Q ≤ ˙ R ∨ ˙ S → ¬ R ∨ ˙ S ∈ A
18 2 3 lnnat ⊢ K ∈ HL ∧ R ∈ A ∧ S ∈ A → R ≠ S ↔ ¬ R ∨ ˙ S ∈ A
19 5 9 10 18 syl3anc ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q ∧ R ∈ A ∧ S ∈ A ∧ P ∨ ˙ Q ≤ ˙ R ∨ ˙ S → R ≠ S ↔ ¬ R ∨ ˙ S ∈ A
20 17 19 mpbird ⊢ K ∈ HL ∧ P ∈ A ∧ Q ∈ A ∧ P ≠ Q ∧ R ∈ A ∧ S ∈ A ∧ P ∨ ˙ Q ≤ ˙ R ∨ ˙ S → R ≠ S